Sunday, January 3, 2010

Problem 412: Triangle, Angles, Congruence

Proposed Problem
Click the figure below to see the complete problem 412 about Triangle, Angles, Congruence.

Problem 412: Triangle, Angles, Congruence.
See also:
Complete Problem 412
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 3:37 PM
Labels: , ,

9 comments:

  1. Peter TranMarch 20, 2012 at 12:19 AM

    Another solution with full detail:
    http://img9.imageshack.us/img9/8220/problem412.png

    From C draw CE//AB and CE=AB ( see attached picture)
    ABEC is a parallelogram and ∠ (ECy)= 2x
    Triangle DEC is isosceles => ∠ (DEC)=1/2∠ (ECy)=x
    Quadrilateral BDCE is cyclic with 2 opposite sides paralleled
    So BDCE is an isosceles trapezoid
    And ∠ (BDA)= ∠ (ECy)=2x => Triangles ABD is isosceles
    In triangles ABD 4x=180-40 => x=35

    ReplyDelete
  2. UnknownJuly 22, 2015 at 10:31 PM

    My solution:
    Draw circle 0 (circumcircle of triangle ABD)
    Let E be the intersection of circle 0 with BC.
    Construct AE and DE
    Triangle BED is isosceles since ∠EBD =∠EAD = x
    So ∠BAE = ∠BDE = x
    Now ∠BDC =40º+2x so ∠EDC=40º+x since ∠BDE= x
    Triangles ABE and EDC are similar by SAS
    So ∠ DCE= x , finally x=35º since x+x+40º+2x=180º => 4x=140º => x=35º

    ReplyDelete
  3. Sumith PeirisDecember 7, 2015 at 4:31 AM

    Let the angle bisector of A meet BC at E and let BE = p

    Now < DAE = < DBE = x so ABED is cyclic

    So (a-p)a = bc......(1)
    Angie bisector theorem
    p/(a-p) = c/b .......(2)

    (1) X (2) gives us

    ap = c^2 hence BC is tangential to Tr ACE at A and so < C = x

    Now we know all the angles of Tr. ABC in terms of x and adding 4x+ 40 = 180 and x = 35

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisDecember 13, 2015 at 7:32 AM

    Or far more simply because ABED is cyclic BE = DE and Tr.s ABE and CDE are congruent SAS and so < C = x and the result follows

    ReplyDelete
  5. APOSTOLIS MANOLOUDISJune 17, 2016 at 2:14 PM

    Problem 412
    Brings BE=//DC then BECD is parallelogram so <BcE=x=< BAE=<BEA.Then BECA isisosceles trapezoid so <ACB=<AEB=x . Then 4x+40=180 so x=35.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  6. AnonymousMay 2, 2020 at 8:18 PM

    It is special case of x = 45-a/4 where a = 40.
    The solution is same as Sumith Peiris.

    ReplyDelete
  7. rv.littlemanOctober 25, 2022 at 9:50 AM

    See the drawing

    Define point E in BC such as ∠DAE=x and ∠BAE=x
    ∠DAE=x and ∠DBE=x => ADEB are concyclic
    ∠BAE=x => ∠BDE=x
    ∠BDE=x and ∠BDE=x =>ΔBED is isosceles in E
    ∠BDC= ∠ABD+ ∠BAD= 40+2x and ∠BDE=x => ∠EDC=40+x
    => ∠EDC= ∠ABE= 40+x
    => ΔABE is congruent to ΔCDE (SAS)
    => EA=EC
    =>ΔAEC is isosceles in E
    => ∠CAE=∠ACE=x
    In ΔCDE : 2x+(40+x)+x= π
    =>4x= π-40=140
    Therefore x=35

    ReplyDelete
  8. AnonymousDecember 9, 2023 at 9:41 PM

    In triangle ABD
    sin(140-2x)/AB=sin2x/BD
    AB/BD=sin(140-2x)/sin2x-------(1)

    In triangle BCD
    sinx/CD=sin(140-3x)/BD
    CD/BD=sinx/sin(140-3x)-------(2)

    As AB=CD, (1)=(2)
    sin(140-2x)/sin2x=sinx/sin(140-3x)
    sinxsin2x=sin(140-2x)sin(140-3x)
    cosx-cos3x=cosx-cos(280-5x)
    cos3x=cos(280-5x)
    x=35

    ReplyDelete

Subscribe to: Post Comments (Atom)